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11.6: Spherical Harmonics

Last updated

Nov 4, 2024
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- 11.5: Now where does the probability peak?
- 11.7: General solutions

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The key issue about three-dimensional motion in a spherical potential is angular momentum. This is true classically as well as in quantum theories. The angular momentum in classical mechanics is defined as the vector (outer) product of $r$ and $p$ ,

$L=r \times p .$

This has an easy quantum analog that can be written as

$\hat{L}=\hat{r} \times \hat{p}$

After exapnsion we find

$\hat{L}=-i \hbar\left(y \frac{\partial}{\partial z}-z \frac{\partial}{\partial y}, z \frac{\partial}{\partial x}-x \frac{\partial}{\partial z}, x \frac{\partial}{\partial y}-y \frac{\partial}{\partial x}\right)$

This operator has some very interesting properties:

$[\hat{L}, \hat{r}]=0 .$

Thus

$[\hat{\boldsymbol{L}}, \hat{H}]=0!$

And even more surprising,

$\left[\hat{L}_x, \hat{L}_y\right]=i \hbar \hat{L}_z .$

Thus the different components of $L$ are not compatible (i.e., can't be determined at the same time). Since $L$ commutes with $H$ we can diagonalise one of the components of $L$ at the same time as $H$ . Actually, we diagonalsie $\hat{L}^2, \hat{L}_z$ and $H$ at the same time!

The solutions to the equation

$\hat{L}^2 Y_{L M}(\theta, \phi)=\hbar^2 L(L+1) Y_{L M}(\theta, \phi)$

are called the spherical harmonics.

Question: check that $\hat{L}^2$ is independent of $r$ !

The label $M$ corresponds to the operator $\hat{L}_z$ ,

$\hat{L}_z Y_{L M}(\theta, \phi)=\hbar M Y_{L M}(\theta, \phi) .$

Question: check that ˆL2\hat{L}^2 is independent of rr !

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Question: check that $\hat{L}^2$ is independent of $r$ !